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Math Help - Quadratic Formula

  1. #1
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    Quadratic Formula

    My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?

    thank you!
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  2. #2
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    Quote Originally Posted by Fibonacci View Post
    My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?

    thank you!
    Hello,

    this is only a guess but...

    If the vertex has the coordinates V(k, h) the the equation of the parabola is:

    y = a(x-k)^2 + h . Equation in turning point form.

    To calculate the intercepts with the x-axis you use that y = 0:

    0 = a(x-k)^2 + h \Longleftrightarrow (x-k)^2 =\frac{-h}{a} \Longleftrightarrow x = k \pm \sqrt{\frac{-h}{a}}

    Maybe this helps a little bit.
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,

    this is only a guess but...

    If the vertex has the coordinates V(k, h) the the equation of the parabola is:

    y = a(x-k)^2 + h . Equation in turning point form.

    To calculate the intercepts with the x-axis you use that y = 0:

    0 = a(x-k)^2 + h \Longleftrightarrow (x-k)^2 =\frac{-h}{a} \Longleftrightarrow x = k \pm \sqrt{\frac{-h}{a}}

    Maybe this helps a little bit.
    yah i think thats right but maybe a specific example if u could....
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  4. #4
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    Quote Originally Posted by Fibonacci View Post
    yah i think thats right but maybe a specific example if u could....
    Hello,

    here comes your example:

    y = -\frac{1}{4} \cdot (x-3)^2 + 4

    The vertex is at V(3, 4) and according to the formula given in my previous post you'll get the zeros Z_1 and Z_2 at:

    x_{1,2} = 3 \pm \sqrt{\frac{-4}{-\frac{1}{4}}} = 3 \pm 4

    I've attached the graph of the parabola so you can check the answer.
    Attached Thumbnails Attached Thumbnails Quadratic Formula-par_vertexform_zeros.gif  
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